Now complex arguments of $\frac{A}{C}x^2+\frac{B}{C}x^1+1+\frac{B}{C}x^{-1}+\frac{A}{C}x^{-2}$ and $\frac{A'}{C'}x^2+\frac{B'}{C'}x^1+1+\frac{B'}{C'}x^{-1}+\frac{A'}{C'}x^{-2}$ are $0$ (for $|x|=1$), so those are real, which leaves argument of $\frac{C'}{C}$ which is also real, and hence $p(x)/q(x)$ also real.

(Actually thinking of $C=0$ brings that $x^{-2}$ works even better than $Cx^{-2}$)

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Observe that $\dfrac{p(z)}{q(z)} = \dfrac{z^{\deg p}p(1/z)}{z^{\deg{q}}q(1/z)}$, so $\dfrac{p(z)q(1/z)}{q(z)p(1/z)} = z^{\deg p - \deg q} = z^0 = 1$. But (if you keep the powers of $z$ with their polynomials) this is a ratio of palindromic polynomials of even degree, which you know works (because products of palindromes are palindromes). Suppose we didn't know that ...

(This is perhaps less surprising if you symmetrize the degrees first and notice that contributions to the arg from one term are cancelled by its palindromic friend. For instance $Az^2 + A = z(A\mathrm{e}^{\mathrm{i} \theta}+A\mathrm{e}^{-\mathrm{i} \theta})$ and $Az^3 + A = z^{3/2}(A\mathrm{e}^{\mathrm{i} (3/2)\theta}+A\mathrm{e}^{-\mathrm{i} (3/2)\theta})$. So the only contribution to the arg of $p$ and $q$ comes from the power of $z$ prefactor, which cancel in the ratio when their degrees are the same.)